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2 years ago

Can someone PLEASE help me with this question?

Mathematics

1 answer:

disa [49]2 years ago

4 0

9514 1404 393

Answer:

Step-by-step explanation:

The future value of the account is given by the formula ...

A = P(1 +r/12)^(12t) . . . . principal P invested at rate r for t years

Solving for t, we find ...

A/P = (1 +r/12)^(12t) . . . . . . . . . . . divide by P

log(A/P) = 12t·log(1 +r/12) . . . . . . take logs

Divide by the coefficient of t, then fill in the numbers.

t = log(A/P)/(12·log(1 +r/12)) = log(202800/93000)/(12·log(1 +.068/12))

t ≈ 11.497

It will take about 11 years for the account balance to reach the desired amount.

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What is the sum of a common geometric series if the first term is 8 and the common ratio is 1/2?

stealth61 [152]

Answer: A

Step-by-step explanation:

The sum to infinity of a geometric series is

S (∞ ) = \frac{a}{1-r} ( - 1 < r < 1 )

where a is the first term 8 and r is the common ratio, hence

S(∞ ) = {8}{1-\{1}{2} } = {8}{1}{2} } = 16

6 0

3 years ago

No wrong answers & NO LINKS , RIGHT answers only please.

vredina [299]

Answer:

Step-by-step explanation:

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4 0

2 years ago

There are 10 employees in a particular division of a company. Their salaries have a mean of 570,000, a median of $55,000,and a s

harkovskaia [24]

Answer:

a) $160,000

b) $55,000

c) $332264.804

Step-by-step explanation:

We are given that there are 10 employees in a particular division of a company and their salaries have a mean of $70,000, a median of $55,000, and a standard deviation of $20,000.

And also the largest number on the list is $100,000 but By accident, this number is changed to $1,000,000.

a) Value of mean after the change in value is given by;

Original Mean = $70,000

$\frac{\sum X}{n}$ = $70,000 ⇒ $\sum X$ = 70,000 * 10 = $700,000

New $\sum X$ after change = $700,000 - $100,000 + $1,000,000 = $1600000

Therefore, New mean = $\frac{1600000}{10}$ = $160,000 .

b) Median will not get affected as median is the middle most value in the data set and since $1,000,000 is considered to be an outlier so median remain unchanged at $55,000 .

c) Original Variance = $20000^{2}$ i.e. $20000^{2}$ = $\frac{\sum x^{2} - n*xbar }{n -1}$

Original $\sum x^{2}$ = ( $20000^{2}$ * (10-1)) + (10 * 70,000) = $3,600,700,000

New $\sum x^{2}$ = $3,600,700,000 - $100,000^{2} + 1,000,000^{2}$ = 9.936007 * $10^{11}$

New Variance = $\frac{new\sum x^{2} - n*new xbar }{n -1}$ = $\frac{9.936007 *10^{11} - 10*160000 }{10 -1}$ = 1.103999 * $10^{11}$ Therefore, standard deviation after change = $\sqrt{1.103999 * 10^{11} }$ = $332264.804 .